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The less-than relation, $<,$ on reals is

1. a partial ordering since it is asymmetric and reflexive
2. a partial ordering since it is antisymmetric and reflexive
3. not a partial ordering because it is not asymmetric and not reflexive
4. not a partial ordering because it is not antisymmetric and reflexive
5. none of the above

To be a POSET ...  It hav to be reflexive,Antisyammetric and transitive .... So option E .. correct me if i am wrong ....

"reals" means set of Real numbers.

Relation $<$ is :

• Irreflexive and hence not reflexive
• Asymmetric and hence antisymmetric and also not symmetric

The relation is not POSET because it is irreflexive.

Condition for Antisymmetric: $\forall a,b \in \mathbb{R}, aRb \neq bRa$ unless $a=b.$

For asymmetric we have a stronger requirement excluding the unless part from the antisymmetric requirement. i.e., $\forall a,b \in \mathbb{R}, aRb \neq bRa$

A relation may be 'not Asymmetric and not reflexive' but still Antisymmetric. Example: $\{(1,1),(1,2)\}$ over the set $\{1,2\}$ which is

• not reflexive because $(2,2)$ is not present
• not irreflexive because $(1,1)$ is present
• not symmetric because $(1,2)$ is present and $(2,1)$ is not
• not asymmetric because $(1,1)$ is present
• but is anti-symmetric.

Antisymmetric and Irreflexive $=$ Asymmetric

Correct Option E.

given relation is not Poset and reason it it is not reflexive..
it has nothing to do with asymmetry.
Sir, please let me know why option 'C' cannot be the answer. I'm confused :(
edited

Definition:A binary relation is a POSET if it is reflexive,antisymmetric and transitive.

Here, "<" is not a POSET because (1,1) is not possible(we can't say 1<1).

By definition of a POSET, we have nothing to do with the asymmetry.

The given relation is not even antisymmetric a<b and b<a cant be true simultaneously even  if a=b. Why u considering it to be antisymmetric
Since < is not antisymmetric and not reflexive then why is option D not the answer.

Oh got it option D means not antisymmetric and "reflexive".....

So yes answer has to be E

This relation is anti-symmetric.

$(a<b)$ will be Anti-symmetric if

$((a < b) \wedge (b < a)) \implies (a=b)$ is $True$.

Note the implication, here, $F \implies <anything>$ is automagically $True$, and hence the relation is Anti symmetric.

Antisymmetric and Irreflexive = Asymmetric

Can anybody explain this??

Asymmetric don't allow flipping as well as same elements.

so here antisymmetric don't allows flipping and irreflexive don't allow same elements thus combining both we get asymmetric.

Since  "<" relation neither reflexive nor patialy ordering on set of real number.

But it is anti Symmetric relation.Therefor Option E will be appropriate option for it.

Can you explain how it is anti symmeteric?
A binary relation $R$ is anti symmetric if for two elements $x$ and $y$, $xRy$ is true but $yRx$ is not. The binary relation < is anti symmetric because for $x$ and $y$, if $x<y$ is true, then certainly $y<x$ is not.
The given relation is not even antisymmetric a<b and b<a cant be true simultaneously even  if a=b. Why u considering it to be antisymmetric

definitely the less than relation won't be reflexive bcz a<a is always false for every a belongs to real number.

now for antisymmetric  (a<b & b<a) implies a=b , as we know if a<b then b<a can never be true at once therefore         (a<b & b<a) results false. but false can implies anything therefore it is antisymmetric.

for a relation to be POSET: it must be reflexive, antisymmetric and transitive

here it is not reflexive but antisymmetric in nature therefore  only E option is correct.